Last updated on July 15th, 2022 at 09:25 am

In this post, we will analyze different scenarios of the *motion of a charged particle in an electric field between two parallel plates*.

## Effect of electric fields on the motion of charged particles

A charged particle in an electric field experiences a force and therefore tends to accelerate.

If the particle is stationary the magnitude of the velocity will change (It will have an acceleration)

If the electric field is parallel to the motion of the

moving charged particle, themagnitude of its velocitywill change.An example of the above case is when electrons are accelerated from the cathode towards the anode in a cathode-ray tube.

If the electric field is at right angles to the velocity of the

moving charged particles, thedirection of the motionof the particles will be changed. The path described by the charged particles will be parabolic – the same shape as a projectile in a uniform gravitational field.The component of the velocity perpendicular to the field is unchanged; the component of the velocity parallel to the field increases uniformly due to acceleration.

## The motion of a charged particle in an electric field between two parallel plates

In this specific case, we will consider the electric field between 2 parallel plates – and we will consider both (a) horizontal plates and (2) parallel plates.

## Motion of a charge between horizontal plates

Two parallel plates are horizontal. The top plate is positively charged and the bottom plate is negatively charged.

A charge q of mass m is moving horizontally i.e. parallel to the horizontal plates (with a velocity V_{x}) into the electric field **E** between the above-mentioned plates.

What will happen here?

- There is no force on the charge along the X axis (horizontally). Hence, it will maintain its horizontal velocity V
_{x}along the X axis. - At the same time, the Electric Field E will apply a force F = Eq on the charge (vertically). As a result, it will have a vertical motion as well with an acceleration.

Here, vertical acceleration**a**= F/m = Eq/m - So, along the direction that is parallel to the plates, the velocity of the charge remains constant. And perpendicular to the plates the charge will come under the influence of an electrostatic force and hence it will deflect with an acceleration in the direction of the force.
- Under the influence of horizontal and vertical motion at a time, it can be shown that the charge will follow a parabola path.
- If the charge is a positive charge then it will be deflected to the negatively charged plate (figure 2), and if it is a negative charge (like an electron), then it will deflect to the positively charged plate (figure 4).
- Now, let’s see what will be the equations for the motion of the charge.

### Horizontal motion

The velocity along the X-axis will remain constant, as said earlier. Here, as per the case statement, it is V_{x}. …….. (1)

The distance traversed x in time t along the X-axis: x = V_{x}t ……….. (2)

### Vertical motion

The velocity of the charge at time t along the Y-axis: V_{y}= at [ as along Y-axis the initial velocity is zero] ………….. (3)

The distance traversed y in time t along the Y-axis or the vertical deflection of the charge due to the electric field: y = (1/2) a t^{2} …….. (4)

Vertical acceleration (or Acceleration along Y-axis) = a = F/m = Eq/m ……………….(5)

Now, let’s say the length of each plate is L. Now if you are asked to find out the deflection of the charge at the time of crossing this length L then what to do?

Note that, the charge will have independent motion along X-axis and Y-axis. With the constant velocity along X-axis, it traverses the length L. So we have to first find out how long the charge takes (T) to cross L with horizontal velocity V_{x}. Then we will find out the deflection in this time period T.

Using equation (2) above, L = V_{x}T

=>T = L/V_{x}

Using equation (4) above,

Vertical deflection y=(1/2) a T^{2}

=> y = (1/2) a (L/V_{x})^{2}

Putting the equation of acceleration a, we get

vertical deflection y = (1/2) a (L/V_{x})^{2}

=> y = (1/2) (qE/m) (L/V_{x})^{2}

## Motion of a charge between vertical plates

In this case, the parallel plates are vertical i.e. along the Y-axis.

The charge q of mass m has started moving parallel to these plates, which means it is moving vertically (along the Y-axis) to get inside the electric field of the 2 charged parallel plates (vertical).

**Here are the listed points:**

- There is no force on the charge along the Y-axis (vertically, i.e. horizontal to the plates). Hence, it will maintain its horizontal velocity V
_{y}along the Y axis. - At the same time, the Electric Field E will apply a force F = Eq on the charge (horizontally). As a result, it will have a horizontal motion as well with an acceleration.
- Here, horizontal acceleration a = F/m = Eq/m
- So to summarize, along the direction that is parallel to the plates, the velocity of the charge remains constant. And perpendicular to the plates the charge will come under the influence of an electrostatic force and hence it will deflect with an acceleration in the direction of the force.
- Under the influence of horizontal and vertical motion at a time, it can be shown that the charge will follow a parabola path.
- If the charge is a positive charge then it will be deflected to the negatively charged plate, and if it is a negative charge (like an electron), then it will deflect to the positively charged plate.
- Now, let’s see what will be the equations for the motion of the charge.

### Vertical motion

The velocity along the Y-axis will remain constant, as said earlier. Here, as per the case statement, it is V_{y}. …….. (1)

The distance traversed y in time t along the Y-axis: y = V_{y}t ……….. (2)

### Horizontal motion

The velocity of the charge at time t along the X-axis: V_{x}= at [ as along X-axis the initial velocity is zero] ………….. (3)

The distance traversed x in time t along the X-axis or the horizontal deflection of the charge due to the electric field: x = (1/2) a t^{2} …….. (4)

Horizontal acceleration (or Acceleration along X-axis) = a = F/m = Eq/m ……………….(5)